Thank you for reporting, we will resolve it shortly
Q.
Two equal forces are acting at a point with an angle of $60^{\circ}$ between them. If the resultant force is equal to $40 \sqrt{3} N$, the magnitude of each force is
Motion in a Plane
Solution:
Given : Force $F_{1}=$ force $F_{2}=F$
or $\quad F_{1}=F_{2}=F$
Resultant force $R=40 \sqrt{3} N$
Angle between two forces $F_{1}$ and $F_{2}$ is $\theta=60^{\circ}$
$\therefore \quad R=\sqrt{F_{1}^{2}+F_{2}^{2}+2 F_{1} F_{2} \cos \theta}$
or $40 \sqrt{3}=\sqrt{F^{2}+F^{2}+2 F^{2} \cos 60^{\circ}}=\sqrt{3 F^{2}}$
or $ F=40 N$